以下是你需要做的:

获取Apache HttpClient，这将使您能够发出所需的请求 用它创建一个HttpPost请求，并添加头应用程序/x-www-form-urlencoded 创建一个StringEntity，并将JSON传递给它 执行调用

代码大致如下(你仍然需要调试它并使它工作):

// @Deprecated HttpClient httpClient = new DefaultHttpClient();
HttpClient httpClient = HttpClientBuilder.create().build();
try {
    HttpPost request = new HttpPost("http://yoururl");
    StringEntity params = new StringEntity("details={\"name\":\"xyz\",\"age\":\"20\"} ");
    request.addHeader("content-type", "application/x-www-form-urlencoded");
    request.setEntity(params);
    HttpResponse response = httpClient.execute(request);
} catch (Exception ex) {
} finally {
    // @Deprecated httpClient.getConnectionManager().shutdown(); 
}

以下是你需要做的:

获取Apache HttpClient，这将使您能够发出所需的请求 用它创建一个HttpPost请求，并添加头应用程序/x-www-form-urlencoded 创建一个StringEntity，并将JSON传递给它 执行调用

代码大致如下(你仍然需要调试它并使它工作):

// @Deprecated HttpClient httpClient = new DefaultHttpClient();
HttpClient httpClient = HttpClientBuilder.create().build();
try {
    HttpPost request = new HttpPost("http://yoururl");
    StringEntity params = new StringEntity("details={\"name\":\"xyz\",\"age\":\"20\"} ");
    request.addHeader("content-type", "application/x-www-form-urlencoded");
    request.setEntity(params);
    HttpResponse response = httpClient.execute(request);
} catch (Exception ex) {
} finally {
    // @Deprecated httpClient.getConnectionManager().shutdown(); 
}

试试下面的代码:

HttpClient httpClient = new DefaultHttpClient();

try {
    HttpPost request = new HttpPost("http://yoururl");
    StringEntity params =new StringEntity("details={\"name\":\"myname\",\"age\":\"20\"} ");
    request.addHeader("content-type", "application/json");
    request.addHeader("Accept","application/json");
    request.setEntity(params);
    HttpResponse response = httpClient.execute(request);

    // handle response here...
}catch (Exception ex) {
    // handle exception here
} finally {
    httpClient.getConnectionManager().shutdown();
}

你可以在Apache HTTP中使用以下代码:

String payload = "{\"name\": \"myname\", \"age\": \"20\"}";
post.setEntity(new StringEntity(payload, ContentType.APPLICATION_JSON));

response = client.execute(request);

此外，您还可以创建一个json对象，并像这样在对象中放入字段

HttpPost post = new HttpPost(URL);
JSONObject payload = new JSONObject();
payload.put("name", "myName");
payload.put("age", "20");
post.setEntity(new StringEntity(payload.toString(), ContentType.APPLICATION_JSON));

@momo对Apache HttpClient的回答，4.3.1版本或更高版本。我使用JSON- java构建我的JSON对象:

JSONObject json = new JSONObject();
json.put("someKey", "someValue");    

CloseableHttpClient httpClient = HttpClientBuilder.create().build();

try {
    HttpPost request = new HttpPost("http://yoururl");
    StringEntity params = new StringEntity(json.toString());
    request.addHeader("content-type", "application/json");
    request.setEntity(params);
    httpClient.execute(request);
// handle response here...
} catch (Exception ex) {
    // handle exception here
} finally {
    httpClient.close();
}

使用HttpURLConnection可能是最简单的。

http://www.xyzws.com/Javafaq/how-to-use-httpurlconnection-post-data-to-web-server/139

你将使用JSONObject或其他东西来构造JSON，但不是用来处理网络;你需要序列化它，然后将它传递给一个HttpURLConnection到POST。